Ive been able to write this but i feel it can be sized down and made better your direct ideas will be of much help as I am only abt 2weeks in c programming but hv been working very hard to learn thks take a look

#include<stdio.h>
int main()
{
int initialrabbits,no_months,Kittens[10],Rabbits,count=1,kits,totalkittens;
printf("\nEnter initial no of rabbits:\a ");
scanf("%d",&initialrabbits);

printf("\\nEnter number of months to be projected:\\a ");
scanf("%d",&no\_months);

    for(count=1;count<=no\_months;count++)
    {
        if (count==1){
        Kittens\[0\]=initialrabbits\*8;
        Rabbits=initialrabbits;
        kits=Kittens\[0\];
        totalkittens=Kittens\[0\];}

        else if (count==2){
        Kittens\[1\]=initialrabbits\*8;
        Rabbits=initialrabbits;
        kits=Kittens\[1\];
        totalkittens=Kittens\[0\]+Kittens\[1\];}

        else if (count==3){
        Kittens\[2\]=(Kittens\[0\]+initialrabbits)\*8;
        Rabbits=Kittens\[0\]+initialrabbits;
        kits=Kittens\[2\];
        totalkittens=Kittens\[1\]+Kittens\[2\];}

        else if (count==4){
        Kittens\[3\]=(Kittens\[0\]+Kittens\[1\]+initialrabbits)\*8;
        Rabbits=Kittens\[0\]+Kittens\[1\]+initialrabbits;
        kits=Kittens\[3\];
        totalkittens=Kittens\[2\]+Kittens\[3\];}

        else if (count==5){
        Kittens\[4\]=(Kittens\[0\]+Kittens\[1\]+Kittens\[2\]+initialrabbits)\*8;
        Rabbits=Kittens\[0\]+Kittens\[1\]+Kittens\[2\]+initialrabbits;
        kits=Kittens\[4\];
        totalkittens=Kittens\[3\]+Kittens\[4\];}

        else if (count==6){
        Kittens\[5\]=(Kittens\[0\]+Kittens\[1\]+Kittens\[2\]+Kittens\[3\]+initialrabbits)\*8;
        Rabbits=Kittens\[0\]+Kittens\[1\]+Kittens\[2\]+Kittens\[3\]+initialrabbits;
        kits=Kittens\[5\];
        totalkittens=Kittens\[4\]+Kittens\[5\];}

        else if (count==7){
        Kittens\[6\]=(Kittens\[0\]+Kittens\[1\]+Kittens\[2\]+Kittens\[3\]+Kittens\[4\]+initialrabbits)\*8;
        Rabbits=Kittens\[0\]+Kittens\[1\]+Kittens\[2\]+Kittens\[3\]+Kittens\[4\]+initialrabbits;
        kits=Kittens\[6\];
        totalkittens=Kittens\[5\]+Kittens\[6\];}

        else if (count==8){
        Kittens\[7\]=(Kittens\[0\]+Kittens\[1\]+Kittens\[2\]+Kittens\[3\]+Kittens\[4\]+Kittens\[5\]+initialrabbits)\*8;
        Rabbits=Kittens\[0\]+Kittens\[1\]+Kittens\[2\]+Kittens\[3\]+Kittens\[4\]+Kittens\[5\]+initialrabbit

Farmers want to be able to project business investment needs over a given period of time (in months). During
the program development, you consult a Rabbit specialist who gives you certain facts about rabbits as stated
below.
i) A female rabbit is called doe, a male rabbit is a buck and baby rabbits are kittens.
ii) Female rabbits take two months to mature and start mating. Once they start mating, they produce every
month. On average, a single rabbit produces 8 kittens every time they produce.
Assuming that the framers will only invest in female rabbits and that male rabbits will always be hired during matting
period and their number is negligible. Assume that no rabbit dies on the farm and that they will all grow at the same
rate.
Write a c program that prompts the farmer to input the initial number of rabbits on the farm and the number of
months the farmer wants to sustain the farm. The program should compute and print out the expected number of
rabbits on the farm for each month until the last month.